efficient variable circshift on 3D matrix

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Jona Gladines
Jona Gladines el 2 de Mzo. de 2022
Respondida: Jona Gladines el 2 de Mzo. de 2022
Hello,
I have a working method of circularly shifting every 60 element vector in a 3D matrix A (300x300x60) over its corresponding value in 2D shift matrix B (300x300) which is relatively slow. I hope there is a faster method than the methods I currently have.
The shifting works as follows: If B(1,1) for example is 10, I want to shift A(1, 1, :) over 10 samples. Every value in B can be different.
My first approach was the following:
for i=1:size(B, 1)
for j=1:size(B, 2)
A(i, j, :) = circshift(A(i, j, :), B(i, j));
end
end
which works, but is relatively slow (0.2s). A second approach was to first reshape matrices A and B to 2D and 1D respectively and get rid of the nested for loop.
a = reshape(A, size(A, 1)*size(A, 2), size(A, 3))';
b = reshape(B, size(B, 1)*size(B, 2), 1);
for i = 1:length(b)
a(:, i) = circshift(a(:, i), b(i));
end
A = reshape(a', size(fm2, 1), size(fm2, 2), size(fm2, 3));
Which also works and is already little bit faster (0.1s).
Is there any other method to do this that would be much faster?
Thanks.
  5 comentarios
Jan
Jan el 2 de Mzo. de 2022
Editada: Jan el 2 de Mzo. de 2022
Providing inputs would be very useful. It matters e.g. if the values of B are unique or if there are typically many same values. Optimizing code can exploit such patterns of the input.
For the test data DGM hast provided, this is twice as fast:
s = size(A);
a = reshape(A, [], s(3))';
b = reshape(B, [], 1);
ub = unique(b);
for i = 1:numel(ub)
m = (b == ub(i));
a(:, m) = circshift(a(:, m), ub(i));
end
A = reshape(a', s);
Jona Gladines
Jona Gladines el 2 de Mzo. de 2022
The shift data is part of a 3D recovery method, of which I cannot disclose more information at this point in time. however I've attached a shift matrix from one of the measurements. However since every item that is measured is different, these values wil also be different for every measurement. Sometimes there will be lots of the same values, other times values might differ more.

Iniciar sesión para comentar.

Respuesta aceptada

Jan
Jan el 2 de Mzo. de 2022
In this code:
s = size(A);
a = reshape(A, [], s(3))';
b = reshape(B, [], 1);
ub = unique(b);
for i = 1:numel(ub)
m = (b == ub(i));
a(:, m) = circshift(a(:, m), ub(i));
end
A = reshape(a', s);
40% of the computing time is spent for transposing. So if you store the data directly in a way, which let the operations work on the first dimension, the computing time is reduced also.

Más respuestas (1)

Jona Gladines
Jona Gladines el 2 de Mzo. de 2022
The data is gathered from camera's and can only be transposed at this step, but your proposed method is certainly fast enough for my purpose.
Thanks to everyone who helped me in this quest.
regards,
Jona

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